Find the general solution of the differential equation: $\frac{dy}{dx} = \frac{3e^{2x} + 3e^{4x}}{e^x + e^{-x}}$

$y = e^{3x} + C$

The correct answer is Option (1) → $y = e^{3x} + C$ ##

Given differential equation is

$\frac{dy}{dx} = \frac{3e^{2x} + 3e^{4x}}{e^x + e^{-x}}$

$\Rightarrow \frac{dy}{dx} = \frac{3e^{2x}(1 + e^{2x})}{e^x + \frac{1}{e^x}}$

$\Rightarrow \frac{dy}{dx} = \frac{3e^{2x}(1 + e^{2x})}{(e^{2x} + 1)} \cdot e^x$

$\Rightarrow \frac{dy}{dx} = 3e^{3x}$

$\Rightarrow dy = 3e^{3x} dx$

Integrating both sides, we get

$\int dy = 3 \int e^{3x} dx$

$\Rightarrow y = 3 \frac{e^{3x}}{3} + C$

$y = e^{3x} + C$